A function satisfies dy/dx = e^(2x - y), and y = 0 when x = 0…

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A function satisfies $\dfrac{dy}{dx} = e^{2x - y}$ , and $y = 0$ when $x = 0$ . Write $e^{2x - y} = e^{2x}e^{-y}$ and separate. The solution has the form $e^{y} = \dfrac{e^{2x}}{2} + C$ . Find the constant $C$ , and then find $y$ at $x = 1$ , correct to four decimal places.

Need a reminder? The key ideaShow

A separable equation $\dfrac{dy}{dx} = f(x)g(y)$ becomes $\dfrac{dy}{g(y)} = f(x)\,dx$ . Integrate both sides with one constant, then use the point.

\displaystyle \int \dfrac{dy}{g(y)} = \int f(x)\,dx + C.

The zeros of $g$ give constant solutions.

The worked solution

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A function satisfies dy/dx = e^(2x - y), and y = 0 when x = 0…

The worked solution

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